%I
%S 1,2,5,11,21,6,27,55,10,46,101,23,89,180,9,114,234,3,139,292,16,206,
%T 416,38,291,591,30,355,4,410,845,25,490,986,40,568,1163,35,665,1331,
%U 56,759,18,798,1659,63,966,1956,65,1100,19,1195,2420,74,1400,22,1453
%N Lexicographically first sequence of distinct positive integers in which successive terms differ by distinct triangular numbers.
%C a(1) = 1; for n > 1, a(n) is the smallest unused positive integer such that a(n)  a(n1) is a triangular number (A000217) that has not already been used as the absolute difference between two successive terms.
%C This sequence differs from Recamán's sequence (A005132) in its starting value (1 vs. 0) and the absolute distance between its successive terms a(n)  a(n1) (any unused triangular number vs. n).
%C Conjecture: this sequence is a permutation of the positive integers.
%C Terms that are less than all subsequent terms: a(1)=1, a(2)=2, a(18)=3, a(29)=4, a(157)=7, a(216)=8, a(254)=13, a(1220)=20; a(?)=33 (does not appear in the first 40000 terms).
%H Jon E. Schoenfield, <a href="/A294745/b294745.txt">Table of n, a(n) for n = 1..10000</a>
%e a(1) = 1 since all terms are positive integers and the sequence is lexicographically first.
%e Writing the kth triangular number A000217(k) as T(k):
%e a(2) = 2 because 2 is the smallest unused positive number that differs from a(1)=1 by a triangular number: 2  1 = 1 = T(1);
%e a(3) = 5 because 5 is the smallest unused positive number that differs from a(2)=2 by a triangular number other than 1 (already used): 5  2 = 3 = T(2).
%e Similarly,
%e a(4) = 11: 11  5 = 6 = T(3);
%e a(5) = 21: 21  11 = 10 = T(4);
%e a(6) = 6:  6  21 = 15 = T(5);
%e a(7) = 27: 27  6 = 21 = T(6);
%e a(8) = 55: 55  27 = 28 = T(7);
%e a(9) = 10: 10  55 = 45 = T(9) (T(8) has not yet been used, but 55  45 = 10 gives a smaller unused number than 55  36 = 19).
%o (MAGMA)
%o a:=[1]; TUsed:=[]; for n in [2..57] do tBest:=0; k:=0; while true do k+:=1; T:=(k*(k+1)) div 2; if not (T in TUsed) then t:=a[n1]T; if t lt 1 then break; end if; if not (t in a) then tBest:=t; end if; end if; end while; if tBest eq 0 then k:=0; while true do k+:=1; T:=(k*(k+1)) div 2; if not (T in TUsed) then t:=a[n1]+T; if not (t in a) then tBest:=t; break; end if; end if; end while; end if; a[n]:=tBest; TUsed[#TUsed+1]:=Abs(a[n]a[n1]); end for; a;
%Y Cf. A000217 (triangular numbers), A005132 (Recamán's sequence).
%K nonn
%O 1,2
%A _Jon E. Schoenfield_, Dec 09 2017
